Question

Answer 1

2.1

a, \(sin3x=-\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow sin3x=sin\left(-\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{\pi}{3}+k2\pi\\3x=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{9}+\dfrac{k2\pi}{3}\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

b, \(sin\left(2x-15^o\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(2x-15^o\right)=sin45^o\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-15^o=45^o+k360^o\\2x-15^o=135^o+k360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=30^o+k180^o\\x=75^o+k180^o\end{matrix}\right.\)

Answer 2

2.1

c, \(sin\left(\dfrac{x}{2}+10^o\right)=-\dfrac{1}{2}\)

\(\Leftrightarrow sin\left(\dfrac{x}{2}+10^o\right)=sin\left(-30^o\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}+10^o=-30^o+k360^o\\\dfrac{x}{2}+10^o=210^o+k360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-40^o+k720^o\\x=400^o+k720^o\end{matrix}\right.\)

d, \(sin4x=\dfrac{2}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=arcsin\dfrac{2}{3}+k2\pi\\4x=\pi-arcsin\dfrac{2}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{arcsin\dfrac{2}{3}}{4}+\dfrac{k\pi}{2}\\x=\dfrac{\pi-arcsin\dfrac{2}{3}}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Answer 3

2.2

a, \(cos\left(x+3\right)=\dfrac{1}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=arccos\dfrac{1}{3}+k2\pi\\x+3=\pi-arccos\dfrac{1}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=arccos\dfrac{1}{3}-3+k2\pi\\x=\pi-arccos\dfrac{1}{3}-3+k2\pi\end{matrix}\right.\)

b, \(cos\left(3x-45^o\right)=\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow cos\left(3x-45^o\right)=cos30^o\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-45^o=30^o+k360^o\\3x-45^o=150^o+k360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=25^o+k120^o\\x=65^o+k120^o\end{matrix}\right.\)